Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $x = \dfrac{n^2 - 7n + 10}{-5n^3 - 40n^2 + 100n} \div \dfrac{n + 5}{2n^2 + 20n} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{n^2 - 7n + 10}{-5n^3 - 40n^2 + 100n} \times \dfrac{2n^2 + 20n}{n + 5} $ First factor out any common factors. $x = \dfrac{n^2 - 7n + 10}{-5n(n^2 + 8n - 20)} \times \dfrac{2n(n + 10)}{n + 5} $ Then factor the quadratic expressions. $x = \dfrac {(n - 2)(n - 5)} {-5n(n - 2)(n + 10)} \times \dfrac {2n(n + 10)} {n + 5} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac { (n - 2)(n - 5) \times 2n(n + 10)} { -5n(n - 2)(n + 10) \times (n + 5)} $ $x = \dfrac {2n(n - 2)(n - 5)(n + 10)} {-5n(n - 2)(n + 10)(n + 5)} $ Notice that $(n - 2)$ and $(n + 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {2n\cancel{(n - 2)}(n - 5)(n + 10)} {-5n\cancel{(n - 2)}(n + 10)(n + 5)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $x = \dfrac {2n\cancel{(n - 2)}(n - 5)\cancel{(n + 10)}} {-5n\cancel{(n - 2)}\cancel{(n + 10)}(n + 5)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $x = \dfrac {2n(n - 5)} {-5n(n + 5)} $ $ x = \dfrac{-2(n - 5)}{5(n + 5)}; n \neq 2; n \neq -10 $